How many words can be formed from the letters of the word "AFTER", so that the vowels never comes together.

Answer: Option D

Explanation:

In a group of 6 boys and 4 girls, four children are to be selected such that
at least one boy should be there.
So we can have

(four boys) or (three boys and one girl) or (two boys and two girls) or (one boy and three gils)

This combination question can be solved as

\begin{aligned}
(^{6}{C}_{4}) + (^{6}{C}_{3} * ^{4}{C}_{1}) + \\
+ (^{6}{C}_{2} * ^{4}{C}_{2}) + (^{6}{C}_{1} * ^{4}{C}_{3}) \\

= \left[\dfrac{6 \times 5 }{2 \times 1}\right] + \left[\left(\dfrac{6 \times 5 \times 4 }{3 \times 2 \times 1}\right) \times 4\right] + \\\left[\left(\dfrac{6 \times 5 }{2 \times 1}\right)\left(\dfrac{4 \times 3 }{2 \times 1}\right)\right] + \left[6 \times 4 \right] \\
= 15 + 80 + 90 + 24\\
= 209
\end{aligned}

Answer: Option B

Explanation:

As we can see the letter "E" is twice in given word, so Required Number
\begin{aligned}
= \frac{7!}{2!} \\
= \frac{7*6*5*4*3*2!}{2!} \\
= 2520
\end{aligned}

Answer: Option B

Explanation:

Vowels in the word "CORPORATION" are O,O,A,I,O
Lets make it as CRPRTN(OOAIO)

This has 7 lettes, where R is twice so value = 7!/2!
= 2520

Vowel O is 3 times, so vowels can be arranged = 5!/3!

= 20

Total number of words = 2520 * 20 = 50400

Answer: Option A

Explanation:

\begin{aligned}
^n{P}_n = n! \\
^5{P}_5 = 5*4*3*2*1 \\
= 120

\end{aligned}

Answer: Option A

Explanation:

word SIGNATURE contains total 9 letters.
There are four vowels in this word, I, A, U and E

Make it as, SGNTR(IAUE), consider all vowels as 1 letter for now
So total letter are 6.
6 letters can be arranged in 6! ways = 720 ways
Vowels can be arranged in themselves in 4! ways = 24 ways

Required number of ways = 720*24 = 17280